Apply the seven-step problem-solving strategy for systems of linear equations to a uniform motion problem involving river currents, using two variables for the still-water speed and the current speed and solving the resulting system by elimination.

Problem: A river cruise ship sailed 60 miles downstream in 4 hours and then took 5 hours to travel upstream back to the dock. Find the speed of the ship in still water and the speed of the river current.

1. Read the problem and draw a diagram showing the ship traveling downstream (with the current) for 4 hours and upstream (against the current) for 5 hours, with each leg covering 60 miles.

2. Identify: The ship's speed in still water and the speed of the current.

3. Name: Let $s$ = the speed of the ship in still water (mph) and $c$ = the speed of the current (mph). Because traveling downstream means the current aids the ship, the downstream rate is $s + c$. Traveling upstream means the current opposes the ship, so the upstream rate is $s - c$. Fill in the rate–time–distance table:

	Rate (mph)	Time (hrs)	Distance (miles)
Downstream	$s + c$	$4$	$60$
Upstream	$s - c$	$5$	$60$

4. Translate into a system of equations. Since distance equals rate times time:

$\left\{\begin{array}{l} 4(s + c) = 60 \\ 5(s - c) = 60 \end{array}\right.$

5. Solve by elimination. First, distribute to convert both equations to standard form:

$4s + 4c = 60$ $5s - 5c = 60$

To eliminate $c$, multiply the first equation by $5$ and the second by $4$ so that the $c$-coefficients become opposites:

$20s + 20c = 300$ $20s - 20c = 240$

Add the equations:

$40s = 540$

Divide both sides by $40$: $s = 13.5$.

Substitute $s = 13.5$ into the original first equation to find $c$:

$4(13.5 + c) = 60$

$54 + 4c = 60$

$4c = 6$

$c = 1.5$

6. Check: Downstream rate: $13.5 + 1.5 = 15$ mph, and $15 \times 4 = 60$ miles ✓. Upstream rate: $13.5 - 1.5 = 12$ mph, and $12 \times 5 = 60$ miles ✓.

7. Answer: The ship travels at $13.5$ mph in still water, and the river current flows at $1.5$ mph.

This example demonstrates how the downstream/upstream framework — where the effective rates are $s + c$ and $s - c$ — naturally produces a system of two linear equations in two unknowns when combined with the distance formula $d = rt$. Because both equations contain the same two variables and are readily expressed in standard form after distributing, elimination is a convenient solving method. A required preliminary step is distributing the time factor across the parentheses ($4(s + c)$ becomes $4s + 4c$) to bring the system into standard form before applying the elimination procedure.